Question

A projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 65m/s at an angle of 37 deg. How much time does it take for the projectile to hit the ground?

Answer 1

You must use the kinematical expressions,
\[y=y_0+v_{0y}t+\frac{1}{2}at^2\\
x=x_0+v_{0x}t\]
where,
\[v_{0y}=v_0\sin \theta\\
v_{0x}=v_0\cos \theta\]
where,
\[v_0=65\ m/s; \theta=37^o; x_0=0\ m; y_0=125\ m; a=-9.8\ m/s^2\]
In order to hit the ground, y=0,
\[y=y_0+v_{0y}t+\frac{1}{2}at^2\Rightarrow 0=39.12t-4.9t^2\Rightarrow t=3.99\ s\]
Then,
\[x=x_0+v_{0x}t \Rightarrow x=51.91\cdot3.99=207.13\ m\]

Answer 2

t = 7.09 s, please check

Answer 3

True,
\[t=39.124.9=7.98⇒x=51.91\cdot7.98=414\ \ m\]